= Solution
We prove by induction on $n$ that every submodule $N\subseteq R^n$ is free. The case $n=0$ is immediate. Let $\pi:R^n\to R$ be projection onto the first coordinate. Since $R$ is a principal <ideal> domain,
$$
\pi(N)=dR
$$
for some $d$. If $d=0$, then $N\subseteq R^{n-1}$ and induction applies. Otherwise choose $x\in N$ with $\pi(x)=d$. For every $y\in N$, write $\pi(y)=rd$; then $y-rx\in\ker(\pi|_N)$. Also $Rx\cap\ker(\pi|_N)=0$ because $R$ is a domain and $d\ne0$. Therefore
$$
N=Rx\oplus\ker(\pi|_N).
$$
The kernel is a submodule of $R^{n-1}$ and is free by induction, while $Rx\cong R$. Hence $N$ is free. This is the <submodule theorem for free modules over a principal ideal domain>.
If $P$ is finitely generated and projective, choose a surjection $f:R^n\to P$. Projectivity supplies $h:P\to R^n$ with $f\circ h=\operatorname{id}_P$, so
$$
R^n=\ker f\oplus h(P).
$$
\b[Thus $P\cong h(P)$ is a submodule of the finitely generated free module $R^n$, and the result just proved shows that $P$ is free.]
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